\(\int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [285]

Optimal result

Integrand size = 26, antiderivative size = 266 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {231 i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} d}+\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-231/1024*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^(1/2)/d*2^(1/2)+231/512*I*a/d/(a+I*a*tan(d
*x+c))^(1/2)+231/640*I*a^3/d/(a+I*a*tan(d*x+c))^(5/2)-1/6*I*a^6/d/(a-I*a*tan(d*x+c))^3/(a+I*a*tan(d*x+c))^(5/2
)-11/48*I*a^5/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(5/2)-33/64*I*a^4/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+
c))^(5/2)+77/256*I*a^2/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}-\frac {231 i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} d}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[Cos[c + d*x]^6*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-231*I)/512)*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((231*I)/640)*a^
3)/(d*(a + I*a*Tan[c + d*x])^(5/2)) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(5/2)) -
(((11*I)/48)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(5/2)) - (((33*I)/64)*a^4)/(d*(a - I*a*Ta
n[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2)) + (((77*I)/256)*a^2)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((231*I)/51
2)*a)/(d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^7\right ) \text {Subst}\left (\int \frac {1}{(a-x)^4 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {\left (11 i a^6\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{12 d} \\ & = -\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {\left (33 i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d} \\ & = -\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (231 i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d} \\ & = \frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (231 i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d} \\ & = \frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}-\frac {\left (231 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{512 d} \\ & = \frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}}-\frac {(231 i a) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{1024 d} \\ & = \frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}}-\frac {(231 i a) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{512 d} \\ & = -\frac {231 i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} d}+\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.20 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i a^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},4,-\frac {3}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{40 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Integrate[Cos[c + d*x]^6*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((I/40)*a^3*Hypergeometric2F1[-5/2, 4, -3/2, (1 + I*Tan[c + d*x])/2])/(d*(a + I*a*Tan[c + d*x])^(5/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (214 ) = 428\).

Time = 108.03 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.65

method	result	size
default	\(-\frac {i \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2816 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-256 \left (\cos ^{6}\left (d x +c \right )\right )+3696 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3465 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3465 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sin \left (d x +c \right )-528 \left (\cos ^{4}\left (d x +c \right )\right )+3465 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+6930 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+3465 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )-3465 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )-3465 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-2310 \left (\cos ^{2}\left (d x +c \right )\right )\right )}{15360 d}\)	\(438\)

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/15360*I/d*(a*(1+I*tan(d*x+c)))^(1/2)*(2816*I*cos(d*x+c)^5*sin(d*x+c)-256*cos(d*x+c)^6+3696*I*cos(d*x+c)^3*s
in(d*x+c)+3465*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)
+1))^(1/2))*cos(d*x+c)+3465*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*si
n(d*x+c)-528*cos(d*x+c)^4+3465*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*
x+c)/(cos(d*x+c)+1))^(1/2))+6930*I*cos(d*x+c)*sin(d*x+c)+3465*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d
*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-3465*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ar
ctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-3465*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c
)/(cos(d*x+c)+1))^(1/2))-2310*cos(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.12 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {{\left (3465 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3465 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-40 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 350 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 1645 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1433 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3184 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 464 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{15360 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/15360*(3465*sqrt(1/2)*d*sqrt(-a/d^2)*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(I*d*e^(2*I*d*x + 2*I*c)
+ I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3465*sqrt(1/2)
*d*sqrt(-a/d^2)*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*
d*x + 2*I*c) + 1))*sqrt(-a/d^2) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*(-40*I*e^(12*I*d*x + 12*I*c) - 350*I*e^(10*I*d*x + 10*I*c) - 1645*I*e^(8*I*d*x + 8*I*c) + 1433*I*e^(6*I*d
*x + 6*I*c) + 3184*I*e^(4*I*d*x + 4*I*c) + 464*I*e^(2*I*d*x + 2*I*c) + 48*I))*e^(-5*I*d*x - 5*I*c)/d

Sympy [F]

\[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \cos ^{6}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*cos(c + d*x)**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.86 \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (3465 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3465 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} - 18480 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 30492 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 12672 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 2816 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 1536 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3}}\right )}}{30720 \, a d} \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/30720*I*(3465*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*
a*tan(d*x + c) + a))) + 4*(3465*(I*a*tan(d*x + c) + a)^5*a^2 - 18480*(I*a*tan(d*x + c) + a)^4*a^3 + 30492*(I*a
*tan(d*x + c) + a)^3*a^4 - 12672*(I*a*tan(d*x + c) + a)^2*a^5 - 2816*(I*a*tan(d*x + c) + a)*a^6 - 1536*a^7)/((
I*a*tan(d*x + c) + a)^(11/2) - 6*(I*a*tan(d*x + c) + a)^(9/2)*a + 12*(I*a*tan(d*x + c) + a)^(7/2)*a^2 - 8*(I*a
*tan(d*x + c) + a)^(5/2)*a^3))/(a*d)

Giac [F]

\[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \cos \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*cos(d*x + c)^6, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^6\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(1/2), x)